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I am following this derivation for the covariant derivative of spinors. I have some questions about this derivation:

1. On page 3 they use the fact, that \begin{align*} V^a(x) = \bar{\Psi}(x)\gamma^a\Psi(x) \end{align*} transforms as a vector. Its not specified what exactly $\bar{\Psi}(x)$ means. Is it the conjugate transpose of $\Psi$? It definitely cant be the complex conjugate of the spinor since it has to be a row vector.
2. I also dont get how they come up with the equality in Equation (19). When I use the parallel transport equation above for the spinor I come up with \begin{align*} S(x+dx)-S(x) = -\bar{\Psi}(x)[\Omega_\nu(x)+\bar{\Omega}_\nu(x)]\Psi(x)dx^\nu \end{align*} Where do the gamma matrices rise from?
3. In equation (21) the third equality arises from the definition of the spin connection \begin{align*} V^a(x+dx)=V^a(x)-\omega^a_{b\nu}V^bdx^\nu \end{align*} as they say in the next sentence. If I use this I end up with \begin{align*} = -\omega^a_{b\nu}\bar{\Psi}(x)\gamma^b\Psi(x)dx^\nu \end{align*} and I cant see, why they dont have the spinors in their result. I assume that all of the problems I have resulting due to the same mistake.

I am following this derivation for the covariant derivative of spinors. I have some questions about this derivation:

1. On page 3 they use the fact, that \begin{align*} V^a(x) = \bar{\Psi}(x)\gamma^a\Psi(x) \end{align*} transforms as a vector. Its not specified what exactly $\bar{\Psi}(x)$ means. Is it the conjugate transpose of $\Psi$? It definitely cant be the complex conjugate of the spinor since it has to be a row vector.
2. I also dont get how they come up with the equality in Equation (19). When I use the parallel transport equation above for the spinor I come up with \begin{align*} S(x+dx)-S(x) = -\bar{\Psi}(x)[\Omega_\nu(x)+\bar{\Omega}_\nu(x)]\Psi(x)dx^\nu \end{align*} Where do the gamma matrices rise from?
3. In equation (21) the third equality arises from the definition of the spin connection \begin{align*} V^a(x+dx)=V^a(x)-\omega^a_{b\nu}V^bdx^\nu \end{align*} as they say in the next sentence. If I use this I end up with \begin{align*} = -\omega^a_{b\nu}\bar{\Psi}(x)\gamma^b\Psi(x)dx^\nu \end{align*} and I cant see, why they dont have the spinors in their result. I assume that all of the problems I have resulting due to the same mistake.

Edit:

I found in this post, that $\bar{\Psi}=\Psi^\dagger\gamma^0$. This solves problem 1 and 2. But I still cant solve problem 3.

I am following this derivation for the covariant derivative of spinors. I have some questions about this derivation:

1. On page 3 they use the fact, that \begin{align*} V^a(x) = \bar{\Psi}(x)\gamma^a\Psi(x) \end{align*} transforms as a vector. Its not specified what exactly $\bar{\Psi}(x)$ means. Is it the conjugate transpose of $\Psi$? It definitely cant be the complex conjugate of the spinor since it has to be a row vector.
2. I also dont get how they come up with the equality in Equation (19). When I use the parallel transport equation above for the spinor I come up with \begin{align*} S(x+dx)-S(x) = -\bar{\Psi}(x)[\Omega_\nu(x)+\bar{\Omega}_\nu(x)]\Psi(x)dx^\nu \end{align*} Where do the gamma matrices rise from?

I am following this derivation for the covariant derivative of spinors. I have some questions about this derivation:

1. On page 3 they use the fact, that \begin{align*} V^a(x) = \bar{\Psi}(x)\gamma^a\Psi(x) \end{align*} transforms as a vector. Its not specified what exactly $\bar{\Psi}(x)$ means. Is it the conjugate transpose of $\Psi$? It definitely cant be the complex conjugate of the spinor since it has to be a row vector.
2. I also dont get how they come up with the equality in Equation (19). When I use the parallel transport equation above for the spinor I come up with \begin{align*} S(x+dx)-S(x) = -\bar{\Psi}(x)[\Omega_\nu(x)+\bar{\Omega}_\nu(x)]\Psi(x)dx^\nu \end{align*} Where do the gamma matrices rise from?
3. In equation (21) the third equality arises from the definition of the spin connection \begin{align*} V^a(x+dx)=V^a(x)-\omega^a_{b\nu}V^bdx^\nu \end{align*} as they say in the next sentence. If I use this I end up with \begin{align*} = -\omega^a_{b\nu}\bar{\Psi}(x)\gamma^b\Psi(x)dx^\nu \end{align*} and I cant see, why they dont have the spinors in their result. I assume that all of the problems I have resulting due to the same mistake.

I am following this derivation for the covariant derivative of spinors. I have some questions about this derivation:

1. On page 3 they use the fact, that \begin{align*} V^a(x) = \bar{\Psi}(x)\gamma^a\Psi(x) \end{align*} transforms as a vector. Its not specified what exactly $\bar{\Psi}(x)$ means. Is it the conjugate transpose of $\Psi$? It definitely cant be the complex conjugate of the spinor since it has to be a row vector.
2. I also dont get how they come up with the equality in Equation (19). When I use the parallel transport equation above for the spinor I come up with \begin{align*} S(x+dx)-S(x) = \bar{\Psi}(x)[\Omega_\nu(x)+\bar{\Omega}_\nu(x)]\Psi(x)dx^\nu \end{align*} Where do the gamma matrices rise from?

I am following this derivation for the covariant derivative of spinors. I have some questions about this derivation:

1. On page 3 they use the fact, that \begin{align*} V^a(x) = \bar{\Psi}(x)\gamma^a\Psi(x) \end{align*} transforms as a vector. Its not specified what exactly $\bar{\Psi}(x)$ means. Is it the conjugate transpose of $\Psi$? It definitely cant be the complex conjugate of the spinor since it has to be a row vector.
2. I also dont get how they come up with the equality in Equation (19). When I use the parallel transport equation above for the spinor I come up with \begin{align*} S(x+dx)-S(x) = -\bar{\Psi}(x)[\Omega_\nu(x)+\bar{\Omega}_\nu(x)]\Psi(x)dx^\nu \end{align*} Where do the gamma matrices rise from?